`( sin A + sin 5A + sin 9A)/( cos A + cos 5A + cos 9A)=`

(sin A + sin5A + sin9A) / (cos A + cos5A + cos9A) =

( sin A + sin 3A + sin 5A + sin 7A)/(cos A + cos 3 A + cos 5 A + cos 7A) = tan 4A.

( sin 4 A sin 3A - sin 5A sin 2A+ sin 7A sin 4A )/(cos 2A cos 3A - cos 2A * cos 7A + cos A * cos 10A )=

(sinA + sin3A) / (cosA + cos3A) = tanalpha, (cosA-cos2A) / (sinA + sin2A) = tanbeta, (sinA + sin3A + sin5A + sin7A) / (cosA + cos3A + cos5A + cos7A) = tan gamma , (cosA-cos5A-cos9A + cos13A) / (sinA-sin5A + sin9A-sin13A) = tandelta, then (a) alpha = 4beta (b) gamma = 2alpha (c) gamma = delta (d) none

((sin 7x + sin 5x ) +(sin 9x + sin 3x))/( (cos 7x + cos 5x ) + (cos 9x + cos 3x ))= tan 6x

((sin 7x + sin 5x ) +(sin 9x + sin 3x))/( (cos 7x + cos 5x ) + (cos 9x + cos 3x ))= tan 6x

((sin 7x + sin 5x ) +(sin 9x + sin 3x))/( (cos 7x + cos 5x ) + (cos 9x + cos 3x ))= tan 6x

((sin 7x + sin 5x ) +(sin 9x + sin 3x))/( (cos 7x + cos 5x ) + (cos 9x + cos 3x ))= tan 6x

Prove that: (cos 2A cos 3A -cos 2A cos 7A + cos A cos 10A)/("sin" 4A sin 3A - sin 2A sin 5A + sin 4A sin 7A) = cot 6A cot 5A

(sin A-sin5A+sin9A-sin13A)/(cos A-cos5A-cos9A+cos13A)=cot4A